Question

Find the value of $k$ if
(i) $1^3+2^3+3^3+...+k^3=6084$
(ii) $1^3+2^3+3^3+....+k^3=2025$

Answer 1

(i) In the given series $1^3+2^3+3^3+....+k^3=6084$, $S_k=6084$ represents the sum of cubes of natural numbers upto $k$. Thus, we can write the formula instead of the series and that is:

Sum of cubes of $n$ natural numbers is $S_n={\left[\frac{n(n+1)}{2}\right]}^2$

Substitute $n=k$ and $S_n=6084$ in $S_n={\left[\frac{n(n+1)}{2}\right]}^2$ as follows:

$S_n={\left[\frac{n(n+1)}{2}\right]}^2\Rightarrow 6084={\left[\frac{k(k+1)}{2}\right]}^2\Rightarrow \frac{k(k+1)}{2}=\sqrt{6084}\Rightarrow \frac{k(k+1)}{2}=78$

$\Rightarrow k(k+1)=78\times 2\Rightarrow k^2+k=156\Rightarrow k^2+k-156=0\Rightarrow k^2+13k-12k-156=0$

$\Rightarrow k(k+13)-12(k+13)=0\Rightarrow (k-12)(k+13)=0\Rightarrow k=12,k=-13$

Ignore the negative values of $k$.

Hence $k=12$.

(ii) In the given series $1^3+2^3+3^3+....+k^3=2015$, $S_k=2015$ represents the sum of cubes of natural numbers upto $k$. Thus, we can write the formula instead of the series and that is:

Sum of cubes of $n$ natural numbers is $S_n={\left[\frac{n(n+1)}{2}\right]}^2$

Substitute $n=k$ and $S_n=6084$ in $S_n={\left[\frac{n(n+1)}{2}\right]}^2$ as follows:

$S_n={\left[\frac{n(n+1)}{2}\right]}^2\Rightarrow 2025={\left[\frac{k(k+1)}{2}\right]}^2\Rightarrow \frac{k(k+1)}{2}=\sqrt{2025}\Rightarrow \frac{k(k+1)}{2}=45$

$\Rightarrow k(k+1)=45\times 2\Rightarrow k^2+k=90\Rightarrow k^2+k-90=0\Rightarrow k^2+10k-9k-90=0$

$\Rightarrow k(k+10)-9(k+10)=0\Rightarrow (k-9)(k+10)=0\Rightarrow k=9,k=-10$

Ignore the negative values of $k$.

Hence $k=9$.